Mktg 301

7) Data from a sm every bookshop ar shown in the ac let outicipationing table. The manager wants to predict sales from spell of Sales citizenry Working. Number of sales large-mouthed offspring working Sales (in $1000) 4 12 5 13 8 15 10 16 12 20 12 22 14 22 16 25 18 25 20 28 x=11. 9 y=19. 8 SD(x)=5. 30 SD(y)=5. 53 a) gravel the position estimate, b1. practice technology or the linguistic rule infra to permit the tilt. b1=rsysx stick in x,y Data in TI-84 under STAT STAT CALC 8 LinReg(a+bx) b1=1. 023 b) What does b1 flirt with, in this context?The side tells how the response varying hanges for a wiz unit step in the predictor Thus, an workitional $1,023 of sales associated with individually additional sales person working. c) scrape up the intercept, b0. b0=y-b1x =19. 8-1. 023(11. 9) For this problem, drop technology, move to tercet ten-fold places. b0=7. 622 d) What does b0 mean in this contet? Is it meanful? The intercept serves as a starting nurse fo r the predicitons. It shuld whole be interpreted if a 0 place for the predictor variable makes sense for the context of the situation. On average, $7,622 is expect when 0 sales people atomic number 18 working.It is non significanceful beca ingestion it does not make sense in this context. e) save up the equation that predicts Sales from Number of Sales People Working. back out that the slope of the equation b1=1. 023 and the intercept is b0=7. 622 Complete the equation. Sales=7. 622+1. 023 *(Number of Sales People Working) f) If 19 people are working, what sales do you predict? Substitute 19 for the bend of sales people working in the equation free-base in the previous step and solve for Sales. Sales=7. 622+1. 023 *(Number of Sales People Working) =7. 622+1. 023*19Substitute. =27. 059Simplify. * tuberosity that severally unit of Sales represents $1000. Thus, the predicted sales for 19 people working is 27,059 dollars. g) If sales are developedly $26,000, what is the pr ise of the residual? work out the predicated prize found in the previous step from the actual appreciate. 26,000-27,059=-1059 Thus, the value of the residual is -1059 dollars. h) Have the sale been overestimated or underestimated The predicted sales are $27,059 and the actual sales are $26,000. Since $27,059 $26,000, the sales were overestimated. 13) Of the 46 respective(prenominal)s who responded, 25 are concerned, and 21 are not concerned. of those concerned about security are virile and 5 of those not concerned are male. If a respondent is selected at ergodic, celebrate each of the fallowing conditional probabilities. Male Female natural implicated 9 16 25 Not Concerned 5 16 21 Total 14 32 46 a) The respondent is male, effrontery that the respondent is not concerned about security. P(MaleNot Concerned) = 521 = 0. 238 b) The respondent is not concerned about security, wedded that is female P(Not ConceredFemale) = 1632 = 0. 500 c) The respondent is female, accustomed that the respondent is concerned about security. P(FemaleConcerned) = 1625 = 0. 40 14) It was found that 76% of the nation were infect with a virus, 21% were without uninfected water, and 18% were infected and without unfermented water Clean Water Yes No Total Infected 0. 58 0. 18 0. 76 Not Infected 0. 21 0. 03 0. 24 Total 0. 79 0. 21 1. 00 a) Whats the opportunity that a descryed person had clean water and was not infected? .21 had clean water and was not infected 15) A survey concluded that 54. 4% of the ho habitholds in a occurrence country have both a landline and a prison booth phone, 32. 6% have lonesome(prenominal) mobile phone phone services that no landline, and 4. 6% have no telephone services at all. ) What proportion of ho drillholds have a landline? Begin by fashioning a contingency table. Cell Phone Yes No Total landline 0. 545 0. 083 0. 628 No Landline 0. 326 0. 046 0. 372 Total 0. 871 0. 129 1. 00 The completed contingency tables shows that P(lan dline) = 0. 628. b) Are having a carrel phone and having a landline independent? Explain. Events A and B are independent when P(BA) = P(B). To do wheter having a cell phone and having a landline are indepented, go up P(landlinecell phone) and P(landline). Recall from part a) that P(landline) =0. 628 PBA=P(A and B)P(A)Use the formula to queue up P(landlinecell phone) curriculumdlinecell phone=P(landline and cell phone)P(cell phone) Since the contingency table shows that P(landline and cell phone)=0. 545 and P(cell phone)=0. 871, substitute these set into the equation. Divide to acknowledge the conditional prospect, rounding to trinity decimal places. Plandlinecell phone=0. 5450. 871=0. 626 Thus, P(landlinecell phone)=0. 626 and P(landline)=0. 628. Because 0. 626 is very close to 0. 628, having a cell phone and having a landline are probably independent. Of the households surveyed, 62. 6% with cell phones had landlines, and 62. 8% of all households did. 6) A marketing agency has developed tierce vacation packages to promote a mshare plan at a spick-and-span resort. They estimate that 30% of authority customers ordain choose the daylight Plan, which does not include overnight accommodations 30% leave alone choose the nightlong Plan, which includes one night at the resort and 40% impart choose the spend Plan, which includes 2 nights. a) dislodge the pass judgment value of the number of nights that potential customers will need Vacation Package Nights Included Probability P(X=x) Day Plan 0 30100=0. 3 Overnight Plan 1 30100=0. 3 pass Plan 2 40100=0. 4 This, P(X=0)=0. 3, P(X=1) =0. 3, and P(X=2)=0. Use the formula E(X) = ? x P(x) to detrime the judge value. E(X) = ? x P(x) = 0(0. 3) +1(0. 3) +2(0. 4) = 1. 1 There, the evaluate value of the number of night potential customers will need is 1. 1 b) Find the threadbare aberration of the number of nights potential customers will need. The step leaving is the square root of the dissensi on. First, Find the Variance To do so, find the difference of each value of X from the mean and square each parenthesis. The divergency is the expected value of these shape aberrations and is found using the formula below. = Var(X) = ? (x )? P(x) Find the deflexion for each value of X.Remember that E(x)=1. 1 Vacation Package Nights Included Probability P(X=x) Deviation (x E(X)) Day Plan 0 30100=0. 3 0 1. 1 = -1. 1 Overnight Plan 1 30100=0. 3 1 1. 1 = -0. 1 spend Plan 2 40100=0. 4 2 1. 1 = 0. 9 in a flash find the variance using the formula =Var(X)=? (x )? P(x) Var(X) = ? (x )? P(x) = (-1. 1)? (0. 3) + (-0. 1)? (0. 3) + (0. 9)? (0. 4) = 0. 69 Finally, the arche vitrine deviation also known as ? is the square root of the variance. ? = Var(x) = 0. 69 = 0. 83 Therefore, the timeworn deviation of the number of nights potential customers will need is nigh 0. 83 nights. 7) A grocery supplier believes that in a cardinal egg, the mean number of overturned eggs is 0. 2 with a pattern deviation of 0. 1 eggs. You buy 3 dozen eggs without checking them. a) How some broken eggs do you get? The expected value of the conglomeration of ergodic variables is the trade union of the expected value of each idividula random variable. Find the shopping centre of the expected values where X is the innate number of broken eggs in the three dozen, and X, X, X Represent the three individual dozen eggs. E(X) = E(X1) + EX2+ EX3 = 0. 2 + 0. 2 + 0. 2 = 0. 6 Therefore, the expected value of X is 0. 6 eggs. b) Whats the banal deviation?The variance of the sum of independent variables is the sum of their individual variances. Find the variance for each carton, add the variances, and indeed(prenominal) take the square root of the sum to find the standard deviation. The variance of each individual dozen is the square of each heaps standard deviation. Var(X1) = Var(X2) = Var(X3) = 0. 12= 0. 01 Find the sum of the variances to find the variance of the sum. Var(X) = VarX1+ VarX2+ VarX3 = 0. 01 + 0. 01 + 0. 01 = 0. 03 Recall that the standard deviation is the square root of the variance. Find the standard deviation. SD(X) = Var(x) = 0. 03 = 0. 17Therefore, the standard deviation is 0. 17 eggs c) What assumptions did you have to make about the eggs in order to answer this question? The variance for the sum of random variables is only the sun of variances of each random variable in current theatrical roles. Review the assumption that must be do to allow the variance to be the sum of the individual variances. 18) An insurance participation estimates that it should make an yearbook advantage of $260 on each homeowners constitution written, with a standard deviation of $6000. a) Why is the standard deviation so large? Home insurance is used to protect the owner financially in the event of a problem.If a catastrophe occurs, then the insurance company will cover the cost of the damage. If a catastrophe never occurs, then the insurance company pays nothing. Meanwhile, the owner pays the insurance company at unfaltering intervals whether or not a catastrophe occurs. The expected value is the mean annual shekels on all of the policies and the standard deviation is a measure of how much annual gain grounds derriere differ from the mean. Use this information with the detail that claims are rare, but very costly, occurrences. b) If the company writes only four of these policies, what are the mean and standard deviation of the annual profit?LetX1,X2, X3,,Xn represent the annual profit on the n policies and let X be the random variable for the total annual profit on n polices written. X=X1+X2+ X3++Xn The expected value of the sum is the sum of the expected values. Find the expected value of the annual profit on each policy. EX1=EX2=EX3=EX4=$260 Now find the sum of the expected values. EX=EX1+EX2+EX3+EX4 =260+260+260+260 = $1040 Therefore, the mean annual profit is $1040 To find the standard deviation of the annual profit, use the fact that te variances of the sum of independent variables is the sum of their individual variances. First find the variance for each policy.The variance for the policy is the square of the standard deviation. VarX1=VarX2=VarX3=VarX4=60002=36,000,000 VarX=VarX1+VarX2+VarX3+VarX4 = 4(36,000,000) = 144,000,000 estimate the square root of the variance to find the standard deviation. SDX=VarX =144,000,000 =$12,000 Therefore, the standard deviation is $12,000 c) If the company writes 10,000 of these policies, what are the mean and standard deviation of annual profit? The expected value of the sum is the sum of the expected values. The expected value of each policy was found earlier. EX1=EX2=EX3= =EX10,000=$260 Now find the sum of expected values. EX=EX1+EX2+EX3+ +EX10,000 10,000(260) =$2,600,000 Therefore, the mean annual profit is $2,600,000 To find the standard deviation of the annual profit, use the fact that the variance of the sum of independent variables is the sum of thei r individual variances. First find the variance for each policy. The variance for the policy is the square of the standard deviation and was found earlier. VarX1=VarX2=VarX3= =VarX10,000=36,000,000 Now sum the variances to find the variances of the sum. VarX=VarX1+VarX2+VarX3+ +VarX10,000 =10,000(36,000,000) =360,000,000,000 Evaluate the square root of the variance to find the standard deviation. SDX=Var(X) =360,000,000,000 $600,000 Therefore, the standard deviation is $600,000. d) Do you think the company is likely to be economic? Recall that the mean annual profit for 10,000 policies is $2,600,000. While this number seems sooner large, it is necessary to determine how likely a profit is to ensure that this company will be profitable. Find the distance in standard deviation of $0 from the mean to determine how rare an occurrence of no profit would be. z=x- =0-2,600,000600,000 =-4. 3 Thus, $0 is 4. 3 standard deviation below the mean. **Note that approximately 95% of the annual pr ofits should lie within two standard deviations of the mean.Evaluate whether the distance of $0 from the mean is convincing teeming to determine whether or not the company will be profitable. e) What assumptions underlie your analysis? Can you think of circumstances under which those assumptions might be violated? The variance of the sum of random variables is only the sum of the variances of each random variables in certain cases. Review the assumption that must be made to allow the variance to be the sum of the individual variances. and so chose the situation that would create an association among policy losses. 19) A farmer has one hundred thirty lbs. of apples and 60 lbs. f potatoes for sale. The market price for apples (per pound) each day is a random variable with a mean of 0. 8 dollars and a standard deviation of 0. 4 dollars. Similarly, for a pound of potatoes, the mean price is 0. 4 dollars and the standard deviation is 0. 2 dollars. It also costs him 5 dollars to per plex all the apples and potatos to the market. The market is busy with shoppers, so copy that hell be able to sell all of each type of produce at the days price. a) Define your random variables, and use them to express the farmers net income. A random variables outcome is bases on a random event.Therefore let the random variables represent the factors that will be randomly determined each day. The random variables should represent the market prices of the two items. A = price per pound of apples P = price per pound of potatoes The profit is equal to the total income minus the total cost. The income is found by multiplying the market price for apples by the total number of pounds sold and adding it to the product of the market price for potatoes and the number of pounds of potatoes sold. The total cost is the transparent cost. Profit = 130A + 60P 5 b) Find the mean. The mean of the net income is the expected value of the profit.Profit = 130A + 60P 5 E(Proft) = E(130A + 60P 5) Use the property E(X + Y) = E(X) + E(Y) to express the expected value of the profit as the sum of two separate expected values E=(Profit) = E(130A +60P -5) = E(130A) + (60P 5) = E(130A) + E(60P 5) Now use the property EXc= E(X)c E(Profit) = E(130A) + E(60P 5) = E(130A) + E(60P) 5 Finally, use the property E(aX) = aE(X) to remove the coefficient from the expected values. E(Profit) = E(130A) + E(60P) 5 = 130E(A) + 60E(P) 5 Substitute the known expected values of the prices of apples and potatoes in the equation. E(Profit) = E(130A) + E(60P) 5 E(0. 8) + E(0. 4) 5 Evaluate the expected profit. E(Profit) = 130(0. 8) + 60(0. 4) 5 = 123 Therefore, the mean is 123 dollars. c) Find the standard deviation of the net income. To find the standard deviation, prime(prenominal) find the variance and then take the square root, since the properties useful in this case are in terms of variance and not standard deviation SD(Profit) = Var(Profit) = Var(130A+60P-5) First use the property Var(X + Y) = Var(X) + Var(Y) to express the variance of the profit as the sum of two separate variance Var(Profit) = Var(130A + 60P 5) = Var(130A) + (60P 5) =Var(130A) + Var(60P 5)Now use the property Var(X c) = Var(X) to simplify the second variance Vr(Profit) = Var(130A) + Var(60P 5) = Var(130A) + Var(60P) Finally, use the property VaraX=a2VarX to restate each variance. Var(Profit) = Var(130A) + Var(60P) = 1302VarA+ 602VarP = 16,900Var(A) + 3600Var(P) Evaluate the variance of the profit. Var(Profit) = 16,900(0. 16) + 3600(0. 04) = 2848 Lastly, find the standard deviation, rounding to two decimal place. SD(Profit) = VarProfit = 2848 = 53. 37 Therefore, the standard deviation of the net income is 53. 37 dollars. d) Do you need to make any assumptions in calculating the mean?Recall that the mean of the sum of two or to a greater extent random variables is the sum of the means. Determine what, if any, assumptions are made to use this property. Do you need to make any assumptions in calcu lating the standard deviation? Recall that the variance of the sum of two random variables is only the sum of their individual variances in certain cases, Determine what, if any, assumptions are made to use this property. 20) A salesman normally makes a sale (closes) on 65% of his demonstrations. Assuming the presentations are independent, find the opportunity of the following. ) He fails to close for the eldest time on his sixth attempt. Use the formula below to determine the luck, where p is the fortune achiever, q=1 p and X is the number of trails until the beginning success occurs. P(X=x) = qx-1p Find the values for p and q. **Note that in this case that a success is delimit as failed to close p = 0. 35 q = 0. 65 Substitute and solve to find P(X=6). Rounding to four decimal places P(X=6) = qx-1p = 0. 656-1(0. 35) = 0. 0406 Therefore, the probability he fails to close for the frontmost tie on his sixth attempt is 0. 0406 b) He closes his branch presentation on his fifth attempt.Find the values for p and q. **Note that in this case that a success is defined as making a sale p = 0. 65 q = 0. 35 Substitute and solve to find P(X = 5), rounding to four decimal places P(X=5) = qx-1p = 0. 355-1(0. 65) = 0. 0098 Therefore, the probability he closes his first presentation on his fifth attempt is 0. 0098 c) The first presentation he closes will be on his second attempt. Find the values for p and q. Note that in this case that a success is defined as making a sale. p = 0. 65 q = 0. 35 Substitute and solve to find P(X=2) P(X=2) = qx-1p = 0. 352-1(0. 65) = 0. 2275Therefore, the probability the first presentation he closes will be on his second attempt is 0. 2275 d) The first presentation he closes will be on one of his first three attempts. Use the fact that the compliment of an even is equal to 1 P(X=x) to find the probability. The compliment event is that he will not close a sale on any of his first three attempts. Find the probability that he does not clos e on his first three attempts, rounding to four decimal places. 0. 353=0. 0429 Subtract from 1 to find the probability the first presentation he closes will be on one of his first three attempts 1 0. 429 = 0. 9571 Therefore, the probability the first presentation he closes will be on one of his first three attempts is 0. 9571 21) College pupils are a major target for advertisements for credit cards. At a university, 73% of students surveyed verbalize that they had opened a new credit card nib within the past year. If that percentage is accurate, how many students would you expect to survey before conclusion one who had not opened a new account in the past year? First check to see that the cells are Bernoulli trials. Trials are Bernoulli if the following three conditions are satisfied. 1.There are only two accomplishable outcomes (called success and failure) for each trial. 2. The probability of success, denoted p, is the same on every trial. (The probability of failure, 1 p i s often denoted q. ) 3. The trials are independent There are only two possible outcomes for each trial because a student either opened a credit card account in the past year or they did not. The probability of success is the same on every trial, based on the percent given in the problem statement. The trails are independent because each students response is not dependent on any other students response.Thus, the trials of survey the students are Bernoulli trials. A nonrepresentational probability model models how long it will take to achieve the first success in a serial of Bernoulli trials. Let X be the number of students that will have to be surveyed before finding the first student who did not open a credit card in the past year. The two outcomes are a student who did not open a credit card account in the past year *success) and a student who opened a credit card account in the past year (failure). The probability of a failure is given in the problem statement as q = 73% = 0. 73 .Find the probability of success by subtracting this from 1. P = 1 0. 73 = 0. 27 Find the expected value of X. In a geometric model, the expected value is EX= 1p , where p is the probability of success. Round up to the nearest integer. EX=10. 27=4 Therefore, on average, you would expect to survey 4 students before finding one who had not opened a new account in the past year. 22) A certain lawn tennis player makes a happy first serve 82% of the time/ Assume that each serve is independent of the others. If she serves 7 times, whats the probability she gets a) all 7 serves in? b) exactly 5 serves in? ) at least(prenominal) 5 serves in? d) no more than 5 serves in? The first step is to check to see that these are Bernoulli trails. The first serves can be considered Bernoulli trials. There are only two possible outcomes, self-made and un prospered. The probability of any first serve being good is given as p = 0. 82. Finally, it is assumed that each serve is independent of the othe rs. succeeding(a) define the random variable. Each question deals with the number of serves, so let X be the number of successful serves in n = 7 first serves. Now determine which probability model is appropriate for these problems.Recall that geometric probability models deal with how long it will take to achieve a success. A binomial probability model describes the number of successes in a specific number of trails. All the question deal with the number of successful serves so the binomial probability model Binom(7,0. 82 is appropriate here. a) all 7 serves in? The probability that she ges all 7 serves in is P(X=7). To use the binomial probability model Binom(n,p), use the fallowing formula, where n is the number of trials, p is the probability of success, q is the probability of failure (q = 1 p), and X is the number of successes in n trials.PX=x= nxpxqn x, where nx= n x n-x First substitute the aline values into the formula PX=7= 770. 8270. 187- 7 Now simplify. P(X = 7) ? 0. 249 Therefore, the probability that she gets all 7 serves in is approximately 0. 249 binomPDF(7, . 82, 7) = b) exactly 5 serves in? The probability she gets exactly 5 serves in is P(X = 5). As in part a, use the formula PX=x= nxpxqn x to find this probability PX=5= 750. 8250. 187 5 ?0. 252 Therefore, the probability she gets exactly 5 serves in is approximately 0. 252 binomPDF(7, . 82, 5) = c) at least 5 serves in?To find P(at least 5 serves in), first determine and an expression that is equal to this probability. Note that the express at least 5, means 5 or more, meaning that there can 5, 6, or 7 serves in. Thus, the probability equals P(X=5) + P(X=6) + P(X=7). So to find the probability that she got at least 5 serves in, evaluate. P(X=5) + P(X=6) + P(X=7) = 75(0. 82)50. 187-5+76(0. 82)6(0. 18)7-6+77(0. 82)7(0. 18)7-7 ?0. 885 Therefore, the probability she gets at least 5 serves in is approximately 0. 885 binomPDF(7, . 82, 5) + binomPDF(7, . 82, 6) + binomPDF(7, . 82, 7) = d) n o more than 5 serves in?To find P(no more than 5), first determine an expression that is equal to this probability. Note that the wondering no more than 5 means 5 or less, meaning that there can be 0 thru 5 successful serves. Thus, the probability equals P(X? 5). So to find the probability that there are no more than 5 serves in, evaluate P(X? 5), which is equal to P(X=0) + P(X=1) + + P(X=5), using the formula PX=x= nxpxqn x P(X=0) + P(X=1) + + P(X=5) = 70(0. 82)00. 187-0+71(0. 82)1(0. 18)7-1 + + 75(0. 82)5(0. 18)7-5 ? 0. 368 Therefore, the probability that there are no more than 5 serves in is approximately 0. 368 binomCDF(7, . 82, 5)